Laplace Transform, Finding solution: y′′+4y′+4y=f(t)

[jakejakejake]

y′′+4y′+4y=f(t)
where f(t)=cos(ωt) if 0<t<π and f(t)=0 if t>π?
The initial conditions are y(0) = 0 , y'(0) = 1

I know that f(t)=cos(ωt)−uπ(t)cos(ωt), the heaviside equation.

AND ω is allowed to vary, supposed to find the general solution, i.e. f(t) in terms of ω

I think that after applying Laplace to both sides, I get: (s + 2)² * F(s) - 1 = s / [ s² + w²] e^(-πs) * [s / (s² + ω²)] But I'm still not sure where to go from here...

Thanks in advance!

 

[HallsofIvy]

y′′+4y′+4y=f(t)
where f(t)=cos(ωt) if 0<t<π and f(t)=0 if t>π?
The initial conditions are y(0) = 0 , y'(0) = 1

I know that f(t)=cos(ωt)−uπ(t)cos(ωt), the heaviside equation.

AND ω is allowed to vary, supposed to find the general solution, i.e. f(t) in terms of ω

You have already used f to mean \(\displaystyle cos(\omega t)\). You mean "find y(t)".

I think that after applying Laplace to both sides, I get: (s + 2)² * F(s) - 1 = s / [ s² + w²] e^(-πs) * [s / (s² + ω²)] But I'm still not sure where to go from here...

Thanks in advance!

Solve your equation for F(s) and look up the "inverse Laplace transform" in a table of transforms.